Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 27 / 1

Use the determinant-normalized slash operator and the usual Hecke operator normalization. A normalized eigenform is a nonzero cusp form $f={\sum }_{r\ge 1}{a}_r{q}^r$ with $a_1=1$ and $T_{n}f=\lambda (n)f$ for every $n$. We first establish the operator facts needed here, rather than assume the Hecke multiplication relations.

Integer matrices of determinant $n$ have the determinant-n matrix representatives for Hecke operators ${\gamma }_{a,b,d}=\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $ad=n$ and $0\le b<d$. Integer row operations give these representatives: the Euclidean algorithm reduces the first column to $(a,0)$, and a shear reduces $b$ modulo $d$. Right multiplication by the modular group permutes the left cosets, so $T_{n}f=n^{k/2-1}{\sum }_{ad=n}{\sum }_{b\mathrm{mod}d}f{|}_k{\gamma }_{a,b,d}$ is modular. Rational changes of coordinate preserve cusp vanishing, so it remains a cusp form. Substituting its Fourier expansion of a modular form and summing $e^{2\pi irb/d}$ givesIn particular $a_1(T_{n}f)=a_n(f)$, and the normalization gives $\lambda (n)=a_n(f)$ for every $n$.

For coprime $u,v$, applying this coefficient formula twice amounts to choosing independently a divisor of $\gcd (m,u)$ and one of $\gcd (m,v)$. Their product runs exactly once through the divisors of $\gcd (m,uv)$, with the same factor $d^{k-1}$ and coefficient index. Thus $T_u{T}_v=T_{uv}$. For a prime $p$, put $P=p^{k-1}$ and $\nu =v_p(m)$. The coefficient of $q^m$ in $T_{p^e}f$ isThe coefficient in $T_p{T}_{p^e}f$ is $A_e(pm)+P{A}_e(m/p)$, the second summand being zero if $p\nmid m$. Expanding the two finite sums and matching their endpoints gives $A_{e+1}(m)+P{A}_{e-1}(m)$ for $e\ge 1$. Equality of all coefficients proves $T_p{T}_{p^e}=T_{p^{e+1}}+p^{k-1}{T}_{p^{e-1}}$. Applying these identities to the nonzero eigenvector $f$ yields

For a prime number $p$, the defining finite sum for the Hecke operator simplifies toThe first summand has no $q^1$ term. In the second, summing the $p$th roots of unity selects precisely the terms whose index is divisible by $p$, and its $q^1$ coefficient is $a_p$. Comparing with $T_{p}f=\lambda (p)f$, using $a_1=1$, gives $\boxed{\lambda (p)=a_p}$. This also follows from the general coefficient computation in part (a).

First the eigenvalues of Hecke operators are real. Here is the needed Petersson inner product argument: unfold the finite coset sum defining $T_n$ into its correspondence domain, then change variables by each matrix. The determinant-normalized slash action preserves $y^{k}dxdy/y^2$ in the paired integrand, and transfers a representative to its inverse. Multiplying that inverse by the positive scalar $n$ gives its adjugate without changing the slash action. Adjugation reverses the integer determinant-$n$ correspondence, so the resulting sum is again $T_n$. Consequently $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$. Taking $g=f$ proves reality of $\lambda (n)$, since $⟨f,f⟩>0$.

Fix one prime $p$ and set $x=\lambda (p)/p^{(k-1)/2}$ and $b_e=\lambda (p^e)/p^{e(k-1)/2}$. The Hecke multiplication relations proved in (a) give $b_0=1$, $b_1=x$ and $b_{e+2}=x{b}_{e+1}-b_e$. The distinct-root assumption excludes $x=\pm 2$.

If $|x|<2$, write $x=2\cos \theta$, $0<\theta <\pi$. Solving the linear recurrence gives $b_e=\sin ((e+1)\theta )/\sin \theta$. If $\theta /(2\pi )$ is rational, infinitely many $e$ satisfy $(e+1)\theta \equiv \theta (\mathrm{mod}2\pi )$, hence $b_e=1$. If it is irrational, density of the corresponding irrational rotation gives infinitely many $e$ in the open arc where $\sin ((e+1)\theta )>\sin \theta /2$.

If $|x|>2$, let $\beta ,{\beta }^{-1}$ be the real roots of $X^2-xX+1$, choosing $|\beta |>1$. Then $b_e=({\beta }^{e+1}-{\beta }^{-e-1})/(\beta -{\beta }^{-1})$. Its even-index terms are positive and tend to infinity, whether $\beta$ is positive or negative. Thus the prime-power coefficients of a level-one eigenform are often large and positive, and in either case